On the Grassmann Graph of Linear Codes
Ilaria Cardinali, Luca Giuzzi, Mariusz Kwiatkowski
May 8, 2020
AbstractLet Γ(n, k) be the Grassmann graph formed by the k-dimensional subspaces of a vector space of dimension n over a finite field F and, for t ∈ N \ {0}, let ∆t(n, k)be the subgraph of
Γ(n, k)formed by the set of linear [n, k]-codes having minimum dual distance at least t + 1. We show that if |F| ≥ n
t
then ∆t(n, k)is connected and it is isometrically embedded in
Γ(n, k). This generalizes some results of [3] and [2].
Keywords: Grassmann graph, Linear Codes, Diameter MSC(2010): 51E22, 94B27
1 Introduction
Let V := V (n, F) be a n-dimensional vector space over a field F and for k = 1, . . . , n − 1, denote by Γ(n, k) the k-Grassmann graph of V , that is the graph whose vertices are the k-subspaces of V and where two vertices X, Y are connected by an edge if and only if dim(X ∩ Y ) = k − 1. See [1] for more detail.
It is interesting to see what properties extend from the graph Γ(n, k) to some of its subgraphs. Suppose that B is a given basis of V ; henceforth we will write the coordinates of the vectors in V with respect to B.
A [n, k]-linear code C is just a k-dimensional vector subspace of V . If BC is an ordered basis
of C, a generator matrix for C is the k × n matrix whose rows are the coordinates of the elements of BC with respect to B. Given a [n, k]-linear code C, its dual code is the [n, n − k]-linear code
C⊥ given by
C⊥:= {v ∈ V : ∀c ∈ C, v · c = 0}
where by · we mean the standard symmetric bilinear form on V given by (v1, . . . , vn) · (c1, . . . , cn) = v1c1+ · · · + vncn.
Since the · is non-degenerate, C⊥⊥= C. We say that C has dual minimum distance t + 1 if and
only if the minimum Hamming distance of the dual C⊥ of C is t + 1.
The condition for a [n, k]-linear C having dual minimum distance at least t + 1 can be easily read on any generator matrix of it. Indeed (see, e.g., Proposition 2.7), C has dual minimum distance at least t + 1 if and only if any t-columns of any generator matrix of C are linearly independent.
For t ∈ N \ {0}, let Ct(n, k)be the set of all [n, k]-linear codes with dual minimum distance at
least t + 1 and denote by ∆t(n, k)the subgraph of Γ(n, k) induced by the elements of Ct(n, k),
i.e. the vertex set of ∆t(n, k)is formed by the elements in Ct(n, k)and two vertices X and Y are
adjacent in ∆t(n, k) if and only if dim(X ∩ Y ) = k − 1. We shall call ∆t(n, k)the Grassmann
Note that for t = 1, C1(n, k) is the class of the non-degenerate [n, k]-linear codes and for t = 2,
C2(n, k)is the class of the projective [n, k]-linear codes.
In general, we say that a subgraph is isometrically embedded in a larger graph if there exists a distance-preserving map among them (see also Definition 2.2); see also [6].
In [2] Kwiatkowski and Pankov studied the graph ∆1(n, k)and more recently in [3] Kwiatkowski,
Pankov and Pasini, considered the graph ∆2(n, k) in the case F is a finite field of order q.
The main results of [2] and [3] state that ∆1(n, k)and ∆2(n, k), if q ≥ n, respectively q ≥ n2
, are connected and have the same diameter as the Grassmann graph Γ(n, k).
In this paper we generalize the results of [2] and [3] to the graphs ∆t(n, k) for arbitrary t ≤ k
and arbitrary fields F.
More in detail, our main result is the following
Theorem 1. Let t, k, n be integers such that 1 ≤ t ≤ k ≤ n < ∞. Suppose that F is a field with
|F| ≥ n t
. Then the graph ∆
t(n, k)is connected and isometrically embedded into the k-Grassmann
graph Γ(n, k). Furthermore, the diameter of ∆t(n, k)and Γ(n, k) are the same.
Remark 1.1. The hypotheses in Theorem 1 are sufficient for the graph ∆t(n, k)to be connected
and to be isometrically embedded into Γ(n, k) but in general they are not necessary. We leave to a future work to determine if the graph ∆t(n, k)might be connected also under some weaker
assumptions on q or t and see if there are cases where the embedding is not isometric. We leave also to a future work to generalize these results for F a possibly non-commutative division ring and also to the infinite dimensional cases both for n and for k.
1.1 Structure of the paper
In Section 2 we recall some basic definitions and preliminary results which shall be used in order to prove Theorem 1. Section 3 contains the proof of our main results; in particular, in Subsection 3.1 we shall prove that the graph ∆t(n, k)is connected and isometrically embedded in
Γ(n, k)for q ≥ nt, while in Subsection 3.2 we shall show that the sets C
t(n, k), when not empty,
always contain codes which are at maximum distance in the Grassmann graph Γ(n, k). Finally, in Subsection 3.3 we prove Theorem 1.
2 Preliminaries
As mentioned in the Introduction, F is a field and V := V (n, F) denotes a n-dimensional vector space over F. Let B = (e1, . . . , en)be a given ordered basis of V with respect to which all the
vectors will be written in coordinates. For k and t integers such that 1 ≤ t ≤ k ≤ n − 1, Ct(n, k)
is the class of [n, k]-linear codes having dual minimum distance at least t + 1. More explicitly, Ct(n, k) := {C ⊆ V :dim C = k, d⊥(C) ≥ t + 1}
where d⊥(C) := d
min(C⊥)is the minimum distance of the dual code C⊥ which means that the weight wt(v) of any codeword v = (v1, . . . , vn) ∈ C⊥ is at least t + 1, i.e.
wt(v) := |{i : vi6= 0}| ≥ t + 1.
Clearly, if Ct(n, k) 6= ∅, then necessarily t ≤ k ≤ n.
If t = k and F = Fq, the elements of Ck(n, k) are exactly the maximum distance separable
[n, k]-codes (see e.g. [5]), that is codes whose minimum distance dmin attains the Singleton bound dmin= n − k + 1; see Corollary 2.8.
Remark 2.1. The condition t ≤ k ≤ n is necessary but in general not sufficient to ensure that
Ct(n, k) is not empty. Indeed, if F = Fq, even for arbitrary values of q, it is not straightforward
to determine if Ct(n, k) 6= ∅or characterize the elements of Ct(n, k). For instance the celebrated
MDS conjecture implies Ck(n, k) = ∅ for n > q + 2. On the other hand, if n < q + 1, then by
Lemma 3.16, Ct(n, k) 6= ∅for all t ≤ k ≤ n.
In order to avoid trivial cases, we shall henceforth suppose that the parameters n, k, t and q if F := Fq, have been chosen so that Ct(n, k) 6= ∅; in Lemma 3.16 it shall be shown that under the
assumptions of Theorem 1 this is always true.
We recall from the Introduction that ∆t(n, k) is the subgraph of Γ(n, k) induced by the
elements of Ct(n, k).
Definition 2.2. Let X ∈ ∆t(n, k). We define the connected component ∆Xt (n, k)of X in ∆t(n, k)
as the subgraph of ∆t(n, k) whose vertices are all Y ∈ ∆t(n, k) such that there is a path in
∆t(n, k) joining X and Y . The graph ∆t(n, k)is connected if ∆Xt (n, k) = ∆t(n, k)for some (and,
consequently for all) X ∈ ∆t(n, k).
For any X, Y ∈ Ct(n, k)write d(X, Y ) for the distance between X and Y in the Grassmann
graph Γ(n, k) and dt(X, Y )for the distance between X and Y in ∆t(n, k).If ∆Xt (n, k) 6= ∆Yt(n, k),
that is X and Y are in different connected components of ∆t(n, k) we put dt(X, Y ) = ∞. We
recall that the diameter of a graph is the maximum of the distances among two of its vertices. Since every edge of ∆t(n, k) is an edge of Γ(n, k), it is straightforward to see that dt(X, Y ) ≥
d(X, Y )for all X, Y ∈ ∆t(n, k).
Definition 2.3. We say that ∆t(n, k) is isometrically embedded in Γ(n, k) if for any X, Y ∈
∆t(n, k) we have dt(X, Y ) = d(X, Y ) = k −dim(X ∩ Y ).
Note that if ∆t(n, k) is isometrically embedded in Γ(n, k), then ∆t(n, k) is also connected.
2.1 Some basic results
Definition 2.4. Let (i1, . . . , it) ∈ Nt be a t-uple of integers such that 1 ≤ i1< i2< · · · < it≤ n.
We denote by Ci1...it := ∩
t
j:=1(xij = 0)the (n − t)-dimensional subspace of V obtained as the
intersection of the coordinate hyperplanes of V of equations xij = 0. We shall call Ci1...it the
(i1, . . . , it)-coordinate subspace of V.
The monomial group M(V ) of V consists of all linear transformations of V which map the set of subspaces {he1i, . . . , heni} in itself. It is straightforward to see that M(V ) ∼= F∗o Sn where
odenotes the wreath product and Sn is the symmetric group of order n; see [5, Chapter 8,§5] for
more details.
Definition 2.5. Two [n, k]-linear codes X and Y are equivalent if there exist a monomial
transformation ρ ∈ M(V ) such that X = ρ(Y ).
Suppose X is a [n, k]-linear code with generator matrix GX. If A ∈ GL(k, F) then G0X = AGX
is also a generator matrix for X.
It follows that two [n, k]-linear codes X and Y with generator matrices respectively GX
and GY are equivalent if there exists A ∈ GL(k, F), a permutation matrix P ∈ GL(n, F) and a
diagonal matrix D ∈ GL(n, F) such that
GX= AGY(P D).
Equivalence between linear codes is an equivalence relation and the equivalence class of a code X corresponds to the orbit of X under the action of M(V ) on the k-dimensional subspaces of V .
Also, it can be readily seen that two codes are equivalent if and only if any two of their generator matrices belong to the same orbit under the action of the group PGL(k, F) : (F∗o S
n),
where PGL(k, F) acts on the right of the generator matrix.
With mostly harmless abuse of notation, in the remainder of this paper we shall not distinguish between the action of M(V ) on the codes (regarded as subspaces of V ) and that on the columns of their generator matrices.
Since equivalent codes have the same parameters (in particular they have the same minimum dual distance), we have that Ct(n, k) consists of unions of orbits under the action of M(V ).
For j = 1, . . . , n, let xj
: V → Fq be the jth-coordinate linear functional of V which acts on
the vectors ei, 1 ≤ i ≤ n, of B as xj(ei) =
0 if i 6= j 1 if i = j .
Observe that, for any v ∈ V and j with 1 ≤ j ≤ n, we have that xj(v) is exactly the j-th
component of v with respect to the basis B. So, if X is a [n, k]-linear code and BX= (b1, . . . , bk)
is a given basis of X with respect to which the generator matrix GX is written, then for any
i, j with 1 ≤ i ≤ k and 1 ≤ j ≤ n, then xj(b
i)is exactly the (i, j)- entry in the matrix GX. So,
the j-th column of GX represents the restriction xj|BX of the functional x
j to the basis B X. By
linearity, we can say that the j-th column of GX represents the restriction xj|X of the functional
xj to X. This has the following important consequence.
Lemma 2.6. Let X ⊆ V be a [n, k]-linear code and GX be a generator matrix of X. A set of
coordinate functionals restricted to X is linearly independent if and only if the columns of GX
representing them are linearly independent.
If F := Fq, we shall also use the notation
[m]q :=
qm− 1
q − 1
for the number of 1-dimensional subspaces of an m-dimensional vector space.
The equivalence between (1) and (2) in the following proposition is well known; however, since many results of the present work rely on it, we present a complete proof for the convenience of the reader.
Proposition 2.7. Let X be a [n, k]-linear code and denote by GX a generator matrix of X. The
following are equivalent.
(1) X has minimum dual distance at least t + 1. (2) Any t columns of GX are linearly independent.
(3) For any 1 ≤ i1 < i2 < · · · < it ≤ n we have dim(X ∩ Ci1...it) = k − t where Ci1...it :=
∩t
j:=1(xij = 0) is the (n − t)-dimensional (i1, . . . , it)-coordinate subspace of V.
Proof. The matrix GX is a parity check matrix for the dual code X⊥. Write the columns of GX
as G1, . . . , Gn and let y = (y1, . . . , yn) ∈ X⊥. Then
GXyt= G1y1+ · · · + Gnyn=0. (1)
Assume (1). Then, for any y ∈ X⊥, y 6= 0, we have wt(y) ≥ t + 1. Suppose by contradiction
that there is a set of t-columns of GX which are linearly dependent. To simplify the exposition,
assume without much loss of generality that this set comprises the first t-columns. Then, G1y1+ G2y2+ · · · + Gtyt=0
with at least one entry yi different from 0; so the vector (y1, . . . , yt, 0, . . . , 0) 6=0 is in X⊥ with
wt(y) ≤ t < t + 1. This contradicts (1).
Conversely, assume (2) and take y ∈ X⊥ with wt(y) = d. Then G
XyT = 0. If d = 0, that is
y =0, then there is nothing to prove. If d 6= 0, suppose, again without much loss of generality, that exactly the first d entries y1, . . . , ydof y are non-zero. Then
G1y1+ · · · + Gdyd =0.
In particular, the first d columns of GX must be linearly dependent; by (2) we necessarily have
d > tsince any set of t columns of GX is independent; this implies (1).
We now prove the equivalence between (2) and (3). Suppose that (3) holds. Then, dim(X ∩ Ci1...it) = k − twhich means that the restrictions x
i1|
X, . . . , xit|X of the t coordinate functionals
xi1, . . . , xit of V to X are linearly independent. Then (2) follows from Lemma 2.6.
Conversely, assume (2) holds and suppose by contradiction that (3) is false, that is that there exists a set of indexes i1, . . . , it such that dim(X ∩ Ci1...it) ≥ k − t + 1. Then, for some
j ∈ {1, . . . , t},we have
X ∩ Ci1...ij−1ij+1...it ⊆ X ∩ Cij.
In terms of coordinate functionals this means xij|
X ∈ hxi1|X, . . . xij−1|X, xij+1|X, . . . , xit|Xi.
So xij|
X is a linear combination of the remaining coordinate functionals. In particular, by
Lemma 2.6, this means that the column Gij of any generator matrix GX of X is a linear
combination of the columns Gi1, . . . , Gij−1, Gij+1, . . . , Git. This contradicts (2).
The following is an immediate consequence of Proposition 2.6.
Corollary 2.8. The set C1(n, k)consists of all [n, k]-linear non-degenerate codes; C2(n, k)consists
of all [n, k]-linear projective codes; the set Ck(n, k), if F = Fq, consists of all [n, k]-linear MDS
codes.
Proof. Only the statement about Ck(n, k)needs to be proved as the descriptions of C1(n, k)and
C2(n, k)follow directly from Proposition 2.7. Suppose C ∈ Ck(n, k), i.e. C is a [n, k]-code having
dual minimum distance at least k + 1. Then, by definition of Ck(n, k), C⊥ is a [n, n − k]-code
with minimum distance at least k + 1 = n − (n − k) + 1 and, as such it is a MDS-code. Since the duals of MDS codes are MDS codes, C = C⊥⊥is also MDS.
Conversely, suppose C to be a [n, k]-linear MDS code; then C⊥is also MDS and has minimum
distance k + 1. It follows that C ∈ Ck(n, k).
3 Proof of Theorem 1
We proceed by steps; first, in Section 3.1 we provide a condition for the graph ∆t(n, k) to be
connected and isometrically embedded in Γ(n, k); then in Section 3.2 we show that any class Ct(n, k) contains elements which are at maximum distance in Γ(n, k); finally in Section 3.3 we
complete the proof of Theorem 1.
3.1 The connectedness of the graph
Lemma 3.1. Let X ∈ Ct(n, k) and let H be a hyperplane of X. If y 6∈ X then
dim(hH, yi ∩ C) ≤ k − t + 1
for every (n − t)-dimensional coordinate subspace C of V. Proof. Suppose by contradiction
dim(hH, yi ∩ C) ≥ k − t + 2. Since H ⊆ X we also have dim(hX, yi ∩ C) ≥ k − t + 2.
Any vector in hH, yi can be written in the form x + αy where x ∈ H and α ∈ F. In particular, there are k − t + 2 linearly independent vectors vi in hX, yi ∩ C of the form
vi= xi+ αiy
where xi ∈ H and αi ∈ F. By Gaussian elimination (we remove y), we have at least k − t + 1
vectors in X which are linearly independent and contained in C; so dim(X ∩ C) ≥ k − t + 1, which is a contradiction because X ∈ Ct(n, k)(see Proposition 2.7).
Lemma 3.2. Let S be a vector space of dimension s, H16= H2 be two distinct hyperplanes of S
with fixed bases B1 and B2. Then, there exists a basis B of S contained in B1∪ B2.
Proof. Since H1 6= H2, there exists at least one element b ∈ B2\ H1. Consider B = B1∪ {b}.
This is a linearly independent set consisting of s distinct elements, B ⊆ S and dim(S) = s. It follows that B is a basis of S with B ⊆ B1∪ B2.
Recall that by d(X, Y ) we mean the distance in Γ(n, k) while dt(X, Y )denotes the distance
in ∆t(n, k).
The following definitions are used in the proof of Lemma 3.8.
Definition 3.3. Put {1,...,n}
t := {(i1, . . . , it) ∈ Nt: 1 ≤ i1 < i2 < · · · < it ≤ n}and define as
colors the elements of it. Endow {1,...,n} t
with the natural lexicographic order on the t-uples. Take X, Y ∈ Ct(n, k)with X 6= Y and let H be a hyperplane of X such that X ∩ Y ⊆ H. The
coloration induced by H is the map
ψH: Y /(H ∩ Y ) →
{1, . . . , n} t
∪ {∞}
sending any vector [p] ∈ Y /(H ∩ Y ) to the smallest (in the lexicographic order) color (i1, . . . , it)
such that
dim(hH, pi ∩ Ci1...it)) = k − t + 1
where Ci1...it is the (i1, . . . , it)-coordinate subspace as defined in Definition 2.4. If no such color
exists we put ψH([p]) = ∞.
The function ψH is well defined. Indeed, if b ∈ [a] = a + (H ∩ Y ), then b = a + h for some
h ∈ H ∩ Y and hH, bi = hH, a + hi = hH, ai; so ψH([a]) = ψH([b]).
Henceforth we shall silently represent each element [p] of Y /(X ∩ Y ) by means of its represen-tative element p.
Definition 3.4. Under the same assumptions as in Definition 3.3, we say that a set T ⊆ Y /(H∩Y )
Definition 3.5. Under the same assumptions as in Definition 3.3, we say that a subspace S of
Y /(H ∩ Y )with dim(S) = s is colorable if there exists at least one monochromatic basis of S. If S is colorable, we define the color ψH(S)of S as the minimum color of a basis of S.
In symbols, let
F(S) := {f = (p1, . . . , ps) : f is a basis of S and ψH(p1) = · · · = ψH(ps) 6= ∞}
be the set of monochromatic bases of S. If f ∈ F(S), denote by ψH(f )the color of any element
in f. Hence
Lemma 3.6. The subspace S is colorable if and only if F(S) 6= ∅.
If S is colorable, the color of S is
ψH(S) :=min{ψH(f ) : f ∈ F(S)}.
In other words, S has color c if there are s independent vectors in S all with the same color c and any other set of s independent vectors in S either is not monochromatic or has color c0≥ c.
Note that a colorable subspace S with color c is not, in general, a monochromatic set.
Lemma 3.7. Let X, Y ∈ Ct(n, k) with dim(X ∩ Y ) = k − d ≥ k − t. If F is a field with |F| ≥ nt
then there exists a code Z ∈ Ct(n, k) such that dim(X ∩ Z) = k − 1 and dim(Z ∩ Y ) = k − d + 1.
Proof. We prove that for every hyperplane H of X containing X ∩ Y , there exists z ∈ Y \ (X ∩ Y )
such that Z := hH, zi ∈ Ct(n, k).
By way of contradiction suppose the contrary. Hence there exists a hyperplane H of X with X ∩ Y ⊆ H such that for every z ∈ Y \ (X ∩ Y ), we have hH, zi 6∈ Ct(n, k).
Equivalently, by Proposition 2.7, we suppose that there exists a hyperplane H of X with X ∩ Y ⊆ H such that for every z ∈ Y \ (X ∩ Y ) there exist indexes i1< i2< · · · < itsuch that
dim(hH, zi) ∩ Ci1...it) ≥ k − t + 1.By Lemma 3.1, we have dim(hH, zi) ∩ Ci1...it) = k − t + 1.
Under these assumptions, we will prove the following claim which leads to a contradiction.
Claim 1. There exist indexes i1, . . . , itand linearly independent vectors p1, . . . , pd∈ Y \ (X ∩ Y )
such that [p1], . . . , [pd] are linearly independent in Y /(X ∩ Y ),
dim Ri= k − t + 1and dim(R1+ · · · + Rd) ≥ k − t + d
where we put Hi:= hH, piiand Ri= Hi∩ Ci1...it.
Note that for every i, Ri⊆ hH, Y i.From Claim 1, we have
dim(Y + R1+ · · · + Rd) ≤dim(H + Y ) = k + d − 1 (2)
and
dim(Y ∩ (R1+ · · · + Rd)) =dim(Y ) + dim(R1+ · · · + Rd) −dim(R1+ · · · + Rd+ Y ) ≥
≥ k + (k − t + d) − (k + d − 1) ≥ k − t + 1. (3) Since Ri⊆ Ci1...it for any i, it follows
dim(Ci1...it∩ Y ) ≥dim(Y ∩ (R1+ · · · + Rt)) ≥ k − t + 1, (4)
So, in order to get the thesis, we need to prove Claim 1.
Let S be a subspace of Y /(H ∩ Y ) with dim(S) = s. We show by induction on s that S is colorable. First, by our hypotheses, for any p ∈ S we have ϕH(p) 6= ∞.
Suppose dim(S) = 2. The projective space PG(S) is then a projective line of PG(Y /H ∩ Y )) so there are |F| + 1 points in PG(S). By hypothesis we have |F| ≥ n
t
possible colors (see Definition 3.3). Hence there are at least 2 linearly independent vectors p1 and p2in S such that
ψH(p1) = ψH(p2).Hence, F(S) 6= ∅. By Lemma 3.6, S is colorable.
Suppose now dim(S) > 2. Put s := dim(S). By induction hypothesis, all subspaces S0 of S
with dimension dim(S0) =dim(S) − 1 are colorable, that is they all admit a monochromatic basis.
For any (s − 2)-subspace S00 of S there are |F| + 1 distinct (s − 1)-dimensional subspaces S0 of S
with S00≤ S0 ≤ S. Also PG(S/S00)is a projective line. Since |F| + 1 > n t
, there are at least two of such subspaces, say S1and S2 with S16= S2 (hence hS1, S2i = S) which have the same color
ψH(S1) = ψH(S2).
Let B1 and B2 be bases of respectively S1 and S2 with ψH(B1) = ψH(B2)(=ψH(S1)). By
Lemma 3.2, there is a basis B of S contained in B1∪ B2. So S admits at least one monochromatic
basis and F(S) 6= ∅. By Lemma 3.6, S is colorable.
Hence, it is always possible to determine a monochromatic set of d independent vectors {p1, p2, . . . , pd} of Y such that [p1], . . . , [pd]are independent in Y /(X ∩ Y ). So, we have (recall
that Hi= hH, pii)
dim(H1∩ Ci1...it) =dim(H2∩ Ci1...it) = · · · =dim(Hd∩ Ci1...it) = k − t + 1. (5)
Since H ⊆ X and X ∈ Ct(n, k), we have dim(H ∩ Ci1...it) ≤ k − t. Recalling that by definition
Ri:= Hi∩Ci1...it = hH, pii∩Ci1...it, we have, by (5), that dim(Ri) =dim(H∩Ci1...it)+1 = k−t+1
and dim(H ∩ Ci1...it) = k − t.
In particular, (note that pi6∈ H), it is always possible to find for 1 ≤ i ≤ d an element hi∈ H
and a non-null element αi∈ Fq such that the point αipi+ hi ∈ Ri is such that
αipi+ hi∈ Ci1...it;
up to a scalar multiple we can assume αi= 1for all i.
We now show that dim(R1+ R2+ · · · + Rd) ≥ k − t + d. Suppose the contrary. Then, without
loss of generality, we can assume Rd⊆ R1+ R2+ · · · + Rd−1. In particular
pd+ hd∈ R1+ · · · + Rd−1,
whence
pd = β1p1+ · · · + βd−1pd−1+ h
with h ∈ H a suitable element and βi∈ F for 1 ≤ i ≤ d − 1. So, given that p1, . . . , pd ∈ Y,
pd− (β1p1+ · · · + βd−1pd−1) = h ∈ H ∩ Y = X ∩ Y,
that is
[pd] + [p1] + · · · + [pd−1] = [0]
in Y /(X ∩ Y ). This contradicts the first part (already proved) of Claim 1, since [p1], . . . , [pd]are
linearly independent vectors of Y /(X ∩ Y ). It follows dim(R1+ R2+ · · · + Rd) ≥ k − t + d. So
Claim 1 holds. This completes the proof of the theorem.
Lemma 3.8. Let X, Y ∈ Ct(n, k)with dim(X ∩ Y ) = k − d ≥ k − t. If F := Fq and q ≥ ntthen
there exist [d]q distinct codes Z ∈ Ct(n, k)such that dim(X ∩Z) = k−1 and dim(Z ∩Y ) = k−d+1.
Proof. For any hyperplane H of X containing X ∩ Y it is possible to apply the argument in the
proof of Lemma 3.7. Thus there are at least [d]q distinct codes Z with the required property.
Lemma 3.9. Suppose F is a field with |F| ≥ n
t
. Then for any X ∈ C
t(n, k) and for every
U ⊂ X with dim U < k − t there exists X0 ∈ Ct(n, k − 1)satisfying U ⊂ X0⊂ X.
Proof. A hyperplane H of X is an element of Ct(n, k − 1) if and only if H does not contain
X ∩ Ci1...it for any i1 < · · · < it. Indeed, if Ci1...it ∩ X ⊆ H for some i1 < · · · < it, then
dim(H ∩Ci1...it) ≥dim(X ∩Ci1...it) = k − t > k − t − 1and H 6∈ Ct(n, k − 1). Conversely, suppose
that for any i1< · · · < it, Ci1...it∩X 6⊆ H; then dim(H ∩X∩Ci1...it) =dim(H ∩Ci1...it) = k−1−t
for any choice of the indexes; so H ∈ Ct(n, k − 1).
By Definition of Ci1...it (see Definition 2.4), there exist at most
n t
distinct spaces C
i1...it.
Now we distinguish two cases.
• If F := Fq is a finite field, each of the spaces Ci1...it is contained in [t]q distinct hyperplanes
of X. So the number of hyperplanes containing at least one X ∩ Ci1...it is at most
n t
[t]q.
On the other hand U is contained in [m]q distinct hyperplanes where m = dim(X/U). Since
m > t we have [m]q≥ [t + 1]q = qt+ qt−1+ · · · + q + 1 ≥ n t qt−1+n t qt−2+ · · · +n t + 1 >n t [t]q.
This shows that there is at least one hyperplane X0 of X containing U and none of the
Ci1...it.
• Suppose F is an infinite field and denote by X∗the dual space of X. Then, for every U ⊆ X
with dim U < k − t the set of hyperplanes X0 of X containing U determine a subspace U
of PG(X∗)of vector dimension at least k − (k − t − 1) = t + 1. Since each of the spaces
X ∩ Ci1...it has dimension k − t (because X ∈ Ct(n, k)), the set Hi1...it of hyperplanes
containing Ci1...it is a subspace of PG(X
∗)of vector dimension t. In particular the set of
all hyperplanes of X containing at least one Ci1...it is the union of
n t
subspaces of PG(X∗ ) each of vector dimension t.
Since the field F is infinite, it is impossible for a projective space of vector dimension at least t + 1 to be the union of a finite number of projective spaces of dimension t.
It follows that there is at least one element X0∈ U \ [ (i1...it)∈ {i1,...,in} t Hi1...it
This leads to the same conclusion as in the case in which F is finite. We are now ready to prove the following theorem.
Theorem 3.10. Suppose F is a field with |F| ≥ n t
. Then ∆
t(n, k)is connected and isometrically
embedded into Γ(n, k).
Proof. Take X, Y ∈ Ct(n, k) with X 6= Y . Put dim(X ∩ Y ) := k − d. If k − d ≥ k − t, then
the thesis follows from Lemma 3.7. Suppose now k − d < k − t. By Lemma 3.9 there exists X0 ⊆ X with X ∩ Y ⊆ X0 and X0∈ C
t(n, k − 1). Let Y0 a (k − d) + 1- dimensional subspace of
Y containing X ∩ Y . Put T = hX0, Y0i. Then dim(T ) = k; also dim(T ∩ Ci1...it) ≤dim(X
0∩ C
i1...it) + 1 = k − 1 − t + 1 = k − t
for all 1 ≤ i1 < i2 < · · · < it ≤ n. In particular T ∈ Ct(n, k) and dim(X ∩ T ) = k − 1,
dim(Y ∩ T ) = k − d + 1. By recursively applying this argument we get that ∆t(n, k)is connected.
By construction, the length dt(X, Y ) of the path joining X and Y in ∆t(n, k) is at most
k −dim(X ∩ Y ) := d(X, Y ), i.e. dt(X, Y ) ≤ d(X, Y ). Since dt(X, Y ) ≥ d(X, Y )in general, we
get the thesis.
Remark 3.11. As a consequence of Theorem 3.10,
diam(∆t(n, k)) ≤diam(Γ(n, k))
when |F| ≥ n t
. In the following section we shall show that these two diameters are actually the same.
3.2 Codes at maximum distance
In this section we do not assume any hypothesis on the parameters, apart that they have been chosen so that there exists at least one [n, k]-linear code with dual minimum distance at least t,i.e. Ct(n, k) 6= ∅. Hence, the graph ∆t(n, k) is not assumed to connected. This observation
justifies the following.
Definition 3.12. We say that two codes in X, Y ∈ Ct(n, k) are opposite in ∆t(n, k) if they
belong to the same connected component ∆X
t (n, k) = ∆Yt(n, k) of ∆t(n, k) and dt(X, Y ) =
diam(∆X t (n, k)).
Definition 3.13. We say that two k-dimensional subspaces X, Y ⊆ V are opposite in Γ(n, k) if
dim(X ∩ Y ) = max{2k − n, 0}.
Observe that if 2k ≤ n, being opposite in Γ(n, k) means dim(X + Y ) = 2k (equivalently, X ∩ Y = {0}), while if 2k > n, it means dim(X + Y ) = n.
Lemma 3.14. Suppose t ≤ k ≤ n, Ct(n, k) 6= ∅ and |F| > max{k, n − k}. Then, for any code
C ∈ Ct(n, k) there exists a code D ∈ Ct(n, k) which is equivalent and opposite to C in Γ(n, k).
Proof. Suppose 2k ≤ n and let C ∈ Ct(n, k) with G as generator matrix. By elementary row
operations on G, which leave C invariant, and column operations by means of ρ ∈ M(V ) (see Definition 2.5), we can obtain a generator matrix
G0:= I A B for an equivalent code ρ(C) =: C0 ∈ C
t(n, k). Here I is the k × k identity matrix, A is a k × k
matrix of rank t (since any t columns of a generator matrix of a code in Ct(n, k) are linearly
independent) and B is a k × (n − 2k). Take λ ∈ F \ {0} and consider the matrix G00λ:= λA I B .
Since G00
λis obtained from G0 by applying transformations induced by the monomial group
M(V ), the code C00
λhaving G00λas generator matrix, is equivalent to C0. In particular Cλ00∈ Ct(n, k).
We want to show that it is always possible to choose λ so that C00
λ and C0 are in direct sum
as subspaces of V , that is the matrix
I A B λA I B
has rank 2k. By elementary row operations, subtracting from the second block λ I A B, we see that the rank of the matrix above is the same as the rank of
I A B
0 I − λA2 (I − λA)B
In particular, this rank is definitely 2k if det(I − λA2) 6= 0.
On the other hand det(I − λA2) = 0if and only if λ−1 is an eigenvalue of A2. Since A2is a
k × k matrix, of rank at most t, the number of its eigenvalues is at most t ≤ k < |F|. So, there is at least one λ ∈ F∗ which is a non-null eigenvalue of A. For such a λ, the matrix G00
λ represents a
code C00
λ ∈ Ct(n, k) such that dim(C0∩ Cλ00) = 0, that is C0 and Cλ00are opposite in Γ(n, k).
Suppose now 2k > n. Let C ∈ Ct(n, k) be a code with generator matrix G. Using elementary
row operations on G and a monomial transformation ρ ∈ M(V ), we can obtain a code C0 := ρ(C)
equivalent to C whose generator matrix G0 is in systematic form, i.e
G0 =In−k 0 A1 0 I2k−n A2
,
where A1 and A2are suitable matrices of dimensions respectively (n − k) × (n − k) and (2k − n) ×
(n − k). For λ ∈ F \ {0}, let now G00λ=λA1 0 In−k λA2 I2k−n 0
and C00
λ be the code with generator
matrix G00
λ. The matrix G00λ is obtained by permuting and multiplying some of the columns of the
matrix G by a non-zero scalar λ; as such the code C00
λ generated by G00λis equivalent to C and C0;
thus, C00
λ ∈ Ct(n, k)for all λ 6= 0.
Since 2k > n, the codes C0 and C00
λ are opposite if and only if dim(C0+ Cλ00) = n, that is to
say the rank of the matrix ¯Gλ=
G0
G00λ
is maximum and equal to n. Explicitly, the structure of the matrix ¯Gλis
¯ Gλ= In−k 0 A1 0 I2k−n A2 λA1 0 In−k λA2 I2k−n 0 .
By using column operations we see that rank In−k 0 A1 0 I2k−n A2 λA1 0 In−k λA2 I2k−n 0 ≥rank In−k 0 A1 0 I2k−n A2 λA1 0 In−k =rank In−k 0 A1 0 I2k−n 0 λA1 0 In−k = rank In−k 0 0 0 I2k−n 0 λA1 0 In−k− λA21 .
So, if λ−1is not an eigenvalue of A2
1, we have that rank ( ¯Gλ) = n. As the matrix A21has dimension
(n − k) × (n − k), we have that A2
1 has at most n − k eigenvalues; as |F| > n − k there are some
values of λ such that this rank is maximum; for these values of λ we get that C0, C00
λ ∈ Ct(n, k)
are opposite.
Observe now that for any two codes X, Y ∈ Ct(n, k)and any η ∈ M(V ), we have d(X, Y ) =
d(η(X), η(Y ))in Γ(n, k). Since C0 = ρ(C)for ρ ∈ M(V ), put D = ρ−1(Cλ00),where Cλ00 is the code constructed above. Then,
d(C, D) = d(ρ−1(C0), ρ−1(Cλ00)) = d(C0, Cλ00).
It follows that C and D are codes in Ct(n, k)which are equivalent and opposite in Γ(n, k).
Corollary 3.15. Suppose Ct(n, k) 6= ∅ and ∆t(n, k) to be isometrically embedded into Γ(n, k).
If |F| > max{k, n − k}, then
diam(∆t(n, k)) =diam(Γ(n, k)).
Proof. By Lemma 3.14, there are at least two codes X, Y ∈ Ct(n, k)with d(X, Y ) = diam(Γ(n, k)).
Since ∆t(n, k) is isometrically embedded in Γ(n, k) we also have dt(X, Y ) =diam(Γ(n, k)). On
the other hand, for any X, Y ∈ Ct(n, k),
dt(X, Y ) = d(X, Y ) ≤diam(Γ(n, k)).
It follows diam(∆t(n, k)) =diam(Γ(n, k)).
Note that for q ≥ n t
, the assumptions of Corollary 3.15 hold.
3.3 Proof of Theorem 1
Lemma 3.16. If F is a field with |F|+1 ≥ n then Ct(n, k) 6= ∅for all t and k with 1 ≤ t ≤ k ≤ n.
Proof. Note that for all t we have Ct(n, k) ⊆ Ct−1(n, k). So, in order to get the lemma we just
need to show that Ck(n, k) 6= ∅under our assumptions. It is well known that if n ≤ q + 1 for
F := Fq a finite field or order q, there exist [n, k]–linear MDS codes. Since the dual of an MDS
code is MDS, it is immediate to see that any k-columns of the generator matrix of a [n, k]–MDS code C are independent. It follows that C ∈ Ck(n, k) 6= ∅.
Suppose now F is an arbirary infinite field. There exist at least n distinct elements a1, a2, . . . , an∈
F. Consider the matrix
G := 1 1 . . . 1 a1 a2 . . . an a2 1 a22 . . . a2n ... ... ak−11 ak−12 . . . ak−1 n
Any k × k minor Mi1...ik of G, comprising the columns i1, . . . , ik is a Vandermonde matrix with
determinant
det(Mi1...ik) =
Y
1≤r<s≤k
(ais− air) 6= 0.
In particular, the code C with generator matrix G belongs to Ck(n, k) which is consequently
non-empty.
References
[1] A.E. Brouwer, A.M. Cohen, A. Neumaier, Distance-Regular Graphs, Springer Verlag (1989). [2] M. Kwiatkowski, M. Pankov, On the distance between linear codes, Finite Fields Appl. 39
(2016), 251–263.
[3] M. Kwiatkowski, M. Pankov, A. Pasini, The graphs of projective codes Finite Fields Appl.
54 (2018), 15–29.
[4] M. Kwiatkowski, M. Pankov, Graphs related to 2-dimensional simplex codes, preprint. [5] F.J. MacWilliams, N.J.A. Sloane, The Theory of Error-Correcting Codes, North-Holland
(1983).
[6] E.E. Shult, Points and Lines, Springer-Verlag (2011) Authors’ addresses:
Ilaria Cardinali
Department of Information Engineering and Mathematics University of Siena
Via Roma 56, I-53100, Siena, Italy ilaria.cardinali@unisi.it
Luca Giuzzi
D.I.C.A.T.A.M. (Section of Mathematics) Università di Brescia
Via Branze 53, I-25123, Brescia, Italy luca.giuzzi@unibs.it
Mariusz Kwiatkowski
Department of Logic and Foundations of Computer Science University of Warmia and Mazury in Olsztyn
Sloneczna 54 Street, PL-10710 Olsztyn, Poland mkw@matman.uwm.edu.pl
